Bài tập củng cố: Căn bậc ba SVIP

Hướng dẫn giải:

$\sqrt[3]{0,001 x^{3}}=\sqrt[3]{(0,1 x)^{3}}=0,1 x;$

$\sqrt[3]{-125 a^{12}}=\sqrt[3]{\left(-5 a^{4}\right)^{3}}=-5 a^{4};$

$\sqrt[3]{27 x^{6}}=\sqrt[3]{\left(3 x^{2}\right)^{3}}=3 x^{2};$

$\sqrt[3]{-0,343 a^{3}}=\sqrt[3]{(-0,7 a)^{3}}=-0,7 a;$

Hướng dẫn giải:

a) $x^{3}=2 \Leftrightarrow x=\sqrt[3]{2}$.

b) $27 x^{3}=-81 \Leftrightarrow x^{3}=-3 \Leftrightarrow \sqrt[3]{x^{3}}=\sqrt[3]{-3} \Leftrightarrow x=-\sqrt[3]{3}$.

c) $\dfrac{1}{2} x^{3}=0,004 \Leftrightarrow x^{3}=0,008 \Leftrightarrow \sqrt[3]{x^{3}}=\sqrt[3]{0,008} \Leftrightarrow x=0,2 .$

d) $\sqrt[3]{3 x+1}=4 \Leftrightarrow 3 x+1=4^{3} \Leftrightarrow x=21.$

e) $\sqrt[3]{3-2 x}=-3 \Leftrightarrow 3-2 x=(-3)^{3} \Leftrightarrow x=15.$

f) $\sqrt[3]{x-2}+2=x \Leftrightarrow \sqrt[3]{x-2}=x-2 \Leftrightarrow x-2=(x-2)^{3}.$

$\Leftrightarrow(x-2)\left[(x-2)^{2}-1\right]=0 \Leftrightarrow\left[\begin{array}{l}x-2=1 \\ (x-2)^{2}=1\end{array}\Leftrightarrow\left[\begin{array}{l}x=2 \\ x-2=1 \\ x-2=-1\end{array}\Leftrightarrow\left[\begin{array}{l}x=2 \\ x=3 \\x=1\end{array}\right.\right.\right..$

Hướng dẫn giải:

a) $\sqrt[3]{x} \lt 2 \Leftrightarrow(\sqrt[3]{x})^{3} \lt 2^{3} \Leftrightarrow x \lt 8$.

b) $\sqrt[3]{2 x-1}>-3 \Leftrightarrow(\sqrt[3]{2 x-1})^{3}>(-3)^{3} \Leftrightarrow 2 x-1>-27 \Leftrightarrow x>-13$

c) $\sqrt[3]{2-3 x} \leq 1 \Leftrightarrow(\sqrt[3]{2-3 x})^{3} \leq 1^{3} \Leftrightarrow 2-3 x \leq 1 \Leftrightarrow 1 \leq x$.

d) $\sqrt[3]{3-4 x} \geq 5 \Leftrightarrow(\sqrt[3]{3-4 x})^{3} \geq 5^{3} \Leftrightarrow 3-4 x \geq 125 \Leftrightarrow-\dfrac{61}{2} \geq x$.

Hướng dẫn giải:

Chú ý hằng đẳng thức:

\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right);\)

\(\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right).\)

a) Vì $7+5 \sqrt{2}=(\sqrt{2})^{3}+1+3 \sqrt{2} \cdot 1(\sqrt{2}+1)=(\sqrt{2}+1)^{3}$ nên

$M=\sqrt[3]{(\sqrt{2}+1)^{3}}=\sqrt{2}+1$

b) Vì $6 \sqrt{3}-10=(\sqrt{3})^{3}-1^{3}-3 \sqrt{3} \cdot 1(\sqrt{3}-1)=(\sqrt{3}-1)^{3}$ nên

$N=\sqrt[3]{(\sqrt{3}-1)^{3}}=\sqrt{3}-1$

c) Vì $5 \sqrt{2}-7=(\sqrt{2})^{3}-1^{3}-3 \sqrt{2} .1(\sqrt{2}-1)=(\sqrt{2}-1)^{3}$ nên

$P=\sqrt[3]{(\sqrt{2}-1)^{3}}-3 \sqrt{2}=\sqrt{2}-1-3 \sqrt{2}=-2 \sqrt{2}-1$

d) Vì $6 \sqrt{3}+10=(\sqrt{3})^{3}+1^{3}+3 \sqrt{3} \cdot 1(\sqrt{3}+1)=(\sqrt{3}+1)^{3}$ nên

$Q=\sqrt[3]{(\sqrt{3}+1)^{3}}-5 \sqrt{3}=\sqrt{3}+1-5 \sqrt{3}=-4 \sqrt{3}+1$

Hướng dẫn giải:

Áp dụng hằng đẳng thức:

\(a^3\pm b^3=\left(a\pm b\right)\left(a^2\mp ab+b^2\right)\)

a) $(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)=(\sqrt[3]{2})^{3}-1^{3}=2-1=1$.

b) $(\sqrt[3]{3}+2)(\sqrt[3]{9}-2 \sqrt[3]{3}+4)=(\sqrt[3]{3})^{3}+2^{3}=3+8=11$.

c) $(\sqrt[3]{7}+\sqrt[3]{2})(\sqrt[3]{49}-\sqrt[3]{14}+\sqrt[3]{4})=(\sqrt[3]{7})^{3}+(\sqrt[3]{2})^{3}=7+2=9$.